Question : A telegraph post is bent at a point above the ground. Its top touches the ground at a distance of $8\sqrt{3}$ metres from its foot and makes an angle of $30^{\circ}$ with the horizontal. The height (in metres) of the post is:
Option 1: 12
Option 2: 16
Option 3: 18
Option 4: 24
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Correct Answer: 24
Solution :
Given: A telegraph post top touches the ground at a distance of $8\sqrt{3}$ m from its foot and makes an angle of $30^{\circ}$ with the horizontal.
We know that $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$ and $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$.
From the given figure,
In $ \triangle CDB$, $\tan30^{\circ}=\frac{CB}{BD}$
⇒ $\frac{1}{ \sqrt{3}}=\frac{BC}{8\sqrt{3}}$
⇒ $BC= 8 \ m$
Again, in $ \triangle CDB$, $\sin30^{\circ}=\frac{CB}{DC}$
⇒ $\frac{1}{ 2}=\frac{8}{CD}$
⇒ $CD=AC=16 \ m$
The height of the post is AB i.e. BC + CD $=8+16=24$ m
Hence, the correct answer is 24 m.
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